# Dividing polynomial

• March 22nd 2011, 11:04 AM
Devi09
Dividing polynomial
I'm having problems dividing the polynomial 2x^5 + 2x^3 + x^2 + 1 by x^2 + 1
• March 22nd 2011, 11:19 AM
HallsofIvy
Looks straight forward to me. But I don't know what to tell you without seeing what you did. Did you remember to include "0x" in the divisor and " $0x^4$" and " $0x$" in the dividend? How many times does $x^2$ divide into $2x^5$?
• March 22nd 2011, 11:39 AM
Devi09
It goes in 2x^3 times but I get x^2 as the next quotient part but the correct quotient should be 2x^3 - 1 according to the answer the book gives
• March 22nd 2011, 04:03 PM
Devi09
$x^2+0x+1sqrt(2x^5+0x^4+2x^3+x^2+0x+1)$

This is the subtracting parts: The quotient i got is $2x^3+x^2+x+1$

Part 1 $2x^5+0x^4+2x^3$

Part 2 $x^4+x^2$

$x^4+x^3+x^2$

Part 3 $x^3+0x$

$x^3+x^2+x$

Part 4 $x^2+1$

$x^2+x+1$

Final $x$
• March 23rd 2011, 05:54 AM
HallsofIvy
Yes, $x^2$ divides into $2x^5$ $2x^3$ times. And multiplying the entire divisor by $2x^3$ gives $2x^5+ 2x^3$
Now subtract that from the dividend:
$2x^5+ 2x^3+ x^2+ 1- (2x^5+ 2x^3)= x^2+ 1$.

And now, of course, $x^2+ 1$ divides into that exactly once: the quotient is $2x^3+ 1$.